# Transitive action of a discrete group on a compact space

Let $$G$$ be a discrete countable group acting on a compact, Hausdorff space $$X$$. Assume that the action is transitive. Namely, $$G\cdot x=X$$, for all $$x\in X$$.

Does it follow that $$X$$ is finite?

I thought that the answer should be yes, as $$G\cdot x$$ is then a discrete compact space, and so must be finite. However, I am not sure anymore that I can claim that it is a discrete space. I'd appreciate any help.

• Notice that under these conditions $G/G_x\cong G\cdot x=X$ homeomorphism (see theorem 6.2 in the book "Crossed Products" by Williams). Thus $G/G_x$ is discrete and compact, so finite, and so is $X$. – Shirly Geffen Jun 23 '19 at 15:50
• @ShirlyGeffen $G/G_x$ need not be homeomorphic to $Gx$. The standard map is a continuous bijection only. There are counterexamples when it is not a homeomorphism. It is true however when $G$ is compact for example (not our case). The theorem most likely assumes something more about $G$ or the action. – freakish Jun 23 '19 at 15:53
• @freakish It is a homeomorphism if $G\cdot x$ is locally closed in $X$, which is the case here. – Shirly Geffen Jun 23 '19 at 15:55
• @ShirlyGeffen the theorem also assumes that $X$ is second countable. Which isn't clear for me why that holds. And with that assumption this can be easily proved using Baire category theorem instead. – freakish Jun 23 '19 at 16:08

Generally any topological space is an image of some discrete space. So your conclusion that $$G\cdot x$$ is discrete is not necessarily true, or requires deeper explanation. But in this scenario it works. Not because $$G$$ is discrete but because it is countable. First of all note that $$X$$ is also countable as an image of a countable set.

Assume that $$X$$ is not discrete. Then there is a point $$x_0\in X$$ which is not isolated. Since $$G$$ acts on $$X$$ transitively and $$x\mapsto gx$$ is a homeomorphism then this shows that no point in $$X$$ is isolated. But a compact Hausdorff space without isolated points has to be uncountable. For the proof see here (plus some discussion regarding related set theoretic axioms, for safety I assume ZFC). Contradiction.

Since $$X$$ is discrete and compact then it has to be finite.

Note that the assumption about $$G$$ being discrete is irrelevant.

• Actually $G$ being discrete is irrelevant but is no restriction since one can always consider $G$ with the discrete topology and proceed. – YCor Jun 24 '19 at 8:21

Since $$G$$ is countable and $$G\cdot x=X$$, we have $$X$$ is countable, compact, Hausdorff. So by Sierpinski-Mazurkiewicz theorem (this seems like an overkill, but I can't think of any simplifications at the moment), $$X$$ is homeomorphic to $$\omega^\alpha\cdot n+1$$, where $$\alpha+1$$ is the Cantor-Bendixon rank of $$X$$, and $$n\geq 1$$ is the cardinality of $$X^{(\alpha)}$$.

Now we need to rule out $$\alpha\geq 1$$. For such cases, note that $$G$$ can only map $$\omega^\alpha\cdot m$$, $$m>0$$ to another point $$\omega^\alpha\cdot m'$$, $$m'>0$$, because every neighbourhood of $$\omega^\alpha\cdot m$$ contains a homeomorphic copy of $$\omega^\alpha+1$$. So the action of the homeomorphism group (hence $$G$$) is not transitive unless $$\alpha=0$$.

• Actually the overkill is based on studying isolated points and uses at the first place that a nonempty countable Hausdorff space has an isolated point, which is enough to proceed (as in freakish's answer). – YCor Jun 24 '19 at 8:23